Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $x = \dfrac{z^2 - 10z}{3z - 9} \times \dfrac{z^2 + 5z + 4}{z^3 - 9z^2 - 10z} $
First factor out any common factors. $x = \dfrac{z(z - 10)}{3(z - 3)} \times \dfrac{z^2 + 5z + 4}{z(z^2 - 9z - 10)} $ Then factor the quadratic expressions. $x = \dfrac {z(z - 10)} {3(z - 3)} \times \dfrac {(z + 1)(z + 4)} {z(z + 1)(z - 10)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {z(z - 10) \times (z + 1)(z + 4) } {3(z - 3) \times z(z + 1)(z - 10) } $ $x = \dfrac {z(z + 1)(z + 4)(z - 10)} {3z(z + 1)(z - 10)(z - 3)} $ Notice that $(z + 1)$ and $(z - 10)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {z\cancel{(z + 1)}(z + 4)(z - 10)} {3z\cancel{(z + 1)}(z - 10)(z - 3)} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $x = \dfrac {z\cancel{(z + 1)}(z + 4)\cancel{(z - 10)}} {3z\cancel{(z + 1)}\cancel{(z - 10)}(z - 3)} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $x = \dfrac {z(z + 4)} {3z(z - 3)} $ $ x = \dfrac{z + 4}{3(z - 3)}; z \neq -1; z \neq 10 $